Question

Find the value of cos $(-{89}^o)+cos(-{87}^o)+cos(-{85}^o)+.....+cos\left({85}^o\right)+cos\left({87}^o\right)+cos\left({89}^o\right)$ is equal to:

Answer 1

we have

$\cos (-{89}^o)+\cos (-87)+....+\cos \left({87}^o\right)+\cos \left(89\right)$

$\Rightarrow$ we know that $\cos (-x)=\cos x$

$=2(\cos i+\cos 3^o+....+\cos {89}^o)$

sum of this series is

$\Rightarrow S_n=\frac{\sin \left(n\frac{\beta }{2}\right)}{\sin \left(\frac{\beta }{2}\right)}\left[\cos (\alpha +(n-1\left)\frac{\beta }{2}\right)\right]$

where $n=45\beta =2$ and $\alpha =1$

So $Sn=2\left[\frac{\sin 45}{\sin 1}\cos 45\right]$

$Sn=2\frac{1}{\sin 1}=\frac{1}{\sin 1}$

Hence the sum of this series is

$[Sn-\frac{1}{\sin 1}]$